Question

If 603 mol of octane combusts, what volume of carbon dioxide is produced at 28.0 °C and 0.995 atm?

Answer 1

2C8H18 + 25O2 ----> 16CO2 + 18H2O

2 moles of octane gives 16 moles of CO2

so 1 mole of octane will give = 16/2 = 8 moles of CO2

so 603 moles of octane will give 8 X 603 = 4824 moles of CO2

now use PV = nRT

P = 0.995 atm
V = ? L
n = 4824
R = 0.0821 L atm/K/mole
T = 28 + 273 = 301 K

0.995 X V = 4824 X 0.0821 X 301

V = 4824 X 0.0821 X 301/ 0.995 = 119810.22 L